Question

A natural number is greater than twise its square root by 3.find the number​

Answer 1

Answer:

the natural number is 9

Answer 2

Answer:

$x = 1,9$

Step-by-step explanation:

$x-2\sqrt{x} =3$

$x - 3 = 2\sqrt{x}$

Squaring both sides,

$(x-3)^{2} = (2\sqrt{x} )^{2}$

⇒ $x^{2} +9-6x= 4x$

⇒ $x^{2} -10x+9=0$

⇒ $x^{2} -(9+1)x+9 =0$

⇒ $x^{2} -9x-x+9$

⇒ $x(x-9)-(x-9)$

⇒ $(x-1)(x-9)$